LESSON PLAN

Solving Two-Step Equations

Apothem Team

Grade 8 · Algebra & Patterning

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

The number trick, performed live: "Think of a number. Double it. Add 10. Halve the result. Subtract your original number. Everyone's answer is… 5." Run it twice; collect the gasp.

Then expose the machinery: call the number $n$ — the trick computes $\frac{2n + 10}{2} - n = n + 5 - n = 5$ . Algebra explains magic. Grade 8 equations extend last year's two-step solves to brackets, fractions, and variables on both sides — enough machinery to build tricks, not just fall for them.

Explore

Solving ladder, four rungs with balance-mat backup on request: (1) brackets first: $3(x - 4) = 18$ — distribute or divide-both-sides-by-3, pairs try both and time them; (2) variables both sides: $7x - 5 = 4x + 16$ ; (3) fractions cleared by multiplying through: $\frac{x}{3} + \frac{x}{4} = 14 \to$ multiply by 12: $4x + 3x = 168 \to x = 24$ ; (4) the trap rung — equations with NO solution ( $2x + 6 = 2x + 1$ → $6 = 1$ ??) and ALL solutions ( $2(x+3) = 2x + 6$ → $6 = 6$ ).

Rung 4's debrief: not every equation pins down one number. A false constant (6 = 1) means NO x works; a true constant (6 = 6) means EVERY x works. The equation was never about x at all.

Formalize

Formalize the full pipeline and the three outcome types:

$\text{simplify} \to \text{collect } x \to \text{peel} \qquad \text{outcomes: one solution } | \text{ none } (6=1) | \text{ all } (6=6)$

Fraction-clearing rule of thumb: multiply both sides by the least common denominator FIRST — equations with fractions become equations without them in one legal move. Brackets: distribute early unless dividing both sides is visibly cleaner ( $3(x-4) = 18$ → divide by 3).

Practice

Practice: six solves up the ladder; two story problems (consecutive integers; a perimeter); one no-solution and one identity, classified with reasons; one build-your-own number trick with its algebraic exposé.

Exit ticket: solve $\frac{2(x + 3)}{5} = 4$ , every move labelled. ( $2(x+3) = 20 \to x + 3 = 10 \to x = 7$ .)

Exit ticket

Exit ticket: solve $\frac{2(x + 3)}{5} = 4$ , every move labelled. ( $2(x+3) = 20 \to x + 3 = 10 \to x = 7$ .)

TIP The no-solution/identity cases feel like failure to students ("I lost the x!"). Reframe loudly: the x cancelling IS the finding — the equation's truth doesn't depend on x. Naming the three outcomes prevents the panic.

WORKED EXAMPLES

Example 1 — Full ladder in one problem:

\frac{3(n - 2)}{4} = \frac{n + 6}{2}

Step 1: Clear fractions — multiply both sides by 4: $3(n - 2) = 2(n + 6)$ .

Step 2: Distribute both sides: $3n - 6 = 2n + 12$ .

Step 3: Collect: $3n - 2n = 12 + 6 \to n = 18$ .

Step 4: Substitute into the ORIGINAL (not a middle line): left $\frac{3(16)}{4} = 12$ ; right $\frac{24}{2} = 12$ ✓.

Step 5: Move-count audit: four moves, each named (clear, distribute, collect, done). Naming moves converts "algebra as vibes" into "algebra as checkable procedure" — and procedure is what survives test pressure.

Example 2 — The consecutive-integers classic, then its twist

The problem: three consecutive integers sum to 96. Find them.

Step 1: Name them structurally: $n$ , $n+1$ , $n+2$ .

Step 2: Equation: $3n + 3 = 96 \to 3n = 93 \to n = 31$ . The integers: 31, 32, 33. Check: sum 96 ✓.

Step 3: The twist that tests understanding: "can three consecutive integers sum to 100?" $3n + 3 = 100 \to 3n = 97$ — not divisible by 3; $n$ isn't an integer. NO such integers exist.

Step 4: The structural reason, better than the algebra: three consecutive integers are (middle − 1) + middle + (middle + 1) = 3 × middle — always a multiple of 3. 96 is; 100 isn't. The equation computed the impossibility; the structure EXPLAINS it. Both layers are worth having.

Example 3 — Build the trick: reverse-engineering an identity

The commission: design a number trick that always outputs 12, and prove it with algebra.

Step 1: Start from the target identity: we need an expression in $n$ that simplifies to 12 regardless of $n$ — the $n$ s must cancel.

Step 2: Draft: "Think of a number. Triple it. Add 36. Divide by 3. Subtract your number." Algebra: $\frac{3n + 36}{3} - n = n + 12 - n = 12$ ✓.

Step 3: Field-test on two wildly different inputs: $n = 5$ : $15, 51, 17, 12$ ✓. $n = -10$ : $-30, 6, 2, 12$ ✓ (negatives ride along for free — the identity never asked $n$ to be nice).

Step 4: The design insight: every "mind-reading" trick is an identity — an equation true for ALL values — dressed in patter. Yesterday's rung-4 curiosity (6 = 6) turned out to be showbiz. Students who can BUILD one own identities in a way no drill achieves.

MATERIALS

•Balance mats (for rungs 1–2)

•Ladder cards

•Number-trick template

•Practice set (PDF)

WATCH FOR

!Distributing into only the first bracket term — still the top error, now with stakes.

!Multiplying only SOME terms when clearing fractions:

\frac{x}{3} + 2 = 5

times 3 becoming

x + 2 = 15

. Every term rides the multiplication.

!"No solution" and "x = 0" conflated.

x = 0

is a perfectly good single solution; "no solution" is a different species entirely.